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X −→ Y Be a Continuous Map of Topological Spaces. We Say That F Is A 16. Maps between manifolds Definition 16.1. Let f : X −! Y be a continuous map of topological spaces. We say that f is a local homeomorphism if for every point x 2 X there is an open neighbourhood U of x such that V = f(U) is open and fjU : U −! V is a homeomorphism. We say that f is an unramified cover if for every point y 2 Y we may find an open neighbourhood V of y, such that the inverse image −1 f (V ) is a disjoint union of open subsets U such that fjU : U −! V is a homeomorphism. Every unramified cover is a local homeomorphism. However: Example 16.2. let Y = C and X = C∗ = C−f0g. Then the inclusion of X into Y is a local homeomorphism but it is not an unramified cover. We will need some basic theory of covering spaces. Definition 16.3. Let f : X −! Y be a continuous map of topological spaces. Let Z be any topological space and let g : Z −! Y be any continuous map. A lift of g to X is a continuous map h: Z −! X such that f ◦ h = g. X - f ? Z - Y In general lifts need neither exist nor are they even unique: Example 16.4. Let f : X = C − f0g −! C = Y be the natural inclusion. Then we cannot lift the identity map g : Y −! Y to X. Example 16.5. Let X and Y be the spaces constructed in (13.2.2). Then there is a natural continuous map f : Y −! X (note that we have switched X and Y , both in terms of (13.2.2) and (16.1)). However the identity map g : X −! X is a continuous map which can be lifted in two different ways. However non-uniqueness is quite pathological: Lemma 16.6. Let f : X −! Y be a local homeomorphism of topologi- cal spaces. Let Z be a connected topological space and let g : Z −! Y be a continuous map. Let hi : Z −! X, i = 1, 2 be any two maps which lift g to X. If X is Hausdorff and there is a point z0 2 Z such that h1(z0) = h2(z0) then h1 = h2. 1 Proof. Let E = f z 2 Z j h1(z) = h2(z) g; be the set of points where h1 and h2 are equal. Then z0 2 E so that E is non-empty by assumption. As X is Hausdorff, the diagonal ∆ ⊂ X ×X is closed. Now E is the inverse image of ∆ by the continuous map h1 × h2 : Z −! X × X, so that E is closed as X is Hausdorff. Suppose that z 2 Z. Let x = hi(z). Then we may find an open neighbourhood U of x and V = f(U) of y = f(x) = g(z), such that fjU : U −! V is a homeomorphism. Since hi are continuous, Wi = −1 hi (U) is open. Let W = W1 \ W2. As f ◦ h1 = g = f ◦ h2; and fjU is injective, W ⊂ E. But then E is open and closed, so that E = Z as Z is connected. Lemma 16.7. Let f : X −! Y be an unramified cover of manifolds. Let γ : [0; 1] −! Y be a path and suppose x 2 X is point of X such that f(x) = y = γ(0). Then γ has a unique lift to a map : [0; 1] −! X; such that (0) = x. Proof. For every point y of the image of γ, we may find a connected −1 open neighbourhood V = Vy ⊂ Y of y such that f (V ) is a disjoint union of open subsets all of which are homeomorphic to V via f. Since the image of γ is compact, we may cover the image by finitely many of these open subsets V1;V2;:::;Vk. Let Wj ⊂ [0; 1] be the inverse image of Vj, and let [ Ij = Wi: i≤j Then Ij is an open subset of [0; 1] which is an interval containing 0. We define j : Ij −! X by induction on j. Suppose that we have defined 0 k, k ≤ j. Pick t 2 Wj \ Wj ⊂ Ii \ Wj+1. Let x = j(t) 2 X (if j = 0 then we take x0 = x). Pick U ⊂ X containing x0 such that U −1 is a connected component of f (Vj+1) which maps homeomorphically down to Vj+1. Then we may clearly extend j to j+1. Uniqueness follows from (16.6). Definition 16.8. Let f : X −! Y be a local homeomorphism. We say that f has the lifting property if given any continuous map γ : [0; 1] −! Y and a point x 2 X such that f(x) = y = γ(0), then we may lift γ to : [0; 1] −! X such that (0) = x. 2 Theorem 16.9. Let f : X −! Y be a local homeomorphism of mani- folds. Then f has the lifting property if and only if f is an unramified cover. We have already shown one direction of (16.9). To prove the other direction we need the following basic result: Theorem 16.10. Let f : X −! Y be a local homeomorphism of mani- folds. Suppose that we have a homotopy G: [0; 1]×[0; 1] −! Y between γ0 and γ1. Suppose also that we can lift γs : [0; 1] −! Y , defined by γs(t) = γ(s; t) to s : [0; 1] −! X. If the function σ : [0; 1] −! X given by σ(s) = s(0) is continuous then we can lift G to a continuous map H : [0; 1] × [0; 1] −! X. Proof. The definition of H : [0; 1] × [0; 1] −! X; 2 is clear; given (s; t) 2 [0; 1] , H(s; t) = s(t). The only thing we need to check is that H is continuous. Note that σ(s) = H(s; 0). Given (s; t) 2 [0; 1]2, let x = H(s; t). Since f is a local homeomor- phism, we may find an open subset U = Us;t such that V = f(U) is −1 open and fjU : U −! V is a homeomorphism. Let W = G (V ). Then W = Ws;t is an open neighbourhood of (s; t), such that H(s; t) 2 U and G(W ) = V . By compactness of [0; 1]2, we may find a finite subcover. It follows that we may subdivide [0; 1]2 into finitely many squares in such a way that each square is mapped by G into an open subset V of Y such that there is an open subset U of X with the property that fjU : U −! V is a homeomorphism and there is a point (s; t) belonging to the square such that H(s; t) 2 U. Suppose that the subdivision is given by k, so that we subdivide each [0; 1] into the intervals [i=k; (i + 1)=k]. By an obvious induction, it suffices to prove that Hj[i=k;(i+1)=k]×[j=k;(j+1)=k]; is continuous, given that σ(t) = H(i=k; t) is continuous. Replacing the square [i=k; (i + 1)=k] × [j=k; (j + 1)=k]; by [0; 1]2 and Y by V , we are reduced to proving that H is continuous, given that there is an open subset U of X which maps homeomorphi- 2 cally down to Y and such that there is a single point (s0; t0) 2 [0; 1] such that H(s0; t0) 2 U. Since fjU : U −! Y is a homeomorphism, we may lift γs0 to a map 0 : [0; 1] −! U. By uniqueness it follows that H(s0; t) 2 U for all t 2 [0; 1]. By uniqueness of the lift of γ, where γ(t) = G(0; t), H(0; t) 2 U 3 for all t 2 [0; 1]. Finally by uniqueness of the lift of γs(t), it follows 2 −1 that H(s; t) 2 U for all (s; t) 2 [0; 1] . But then G = (fjU ) ◦ H is certainly continuous. To prove (16.9), we start with a seemingly special case: Proposition 16.11. Let f : X −! Y be a local homeomorphism of connected manifolds. If f has the lifting property and Y is simply connected then f is an isomorphism. Proof. Pick a 2 X and let b = f(a) 2 Y . Pick y 2 Y . Since Y is a connected manifold it is path connected. Pick a path γ : [0; 1] −! Y such that b = γ(0) and y = γ(1). Let : [0; 1] −! X be a lifting of γ and let x = (1). Then f(x) = y. Thus f is surjective. Suppose that x0, x1 2 X, with y = f(x0) = f(x1). Pick paths i : [0; 1] −! X such that a = i(0) and xi = i(1). Let γi = f ◦ i. As Y is simply connected, we may pick a homotopy G between 0 and 1 fixing y. By (16.10) we may lift G to a continuous function H, such that H(0; 0) = x. Since the function (t) = G(0; t) is continuous, the composition f ◦ is constant (equal to y) and the fibre f −1(y) is discrete, it follows that G(0; t) = x is constant. But then x1 = G(0; 1) = G(0; 0) = x0. Thus f is injective. As f is a local homeomorphism, it follows that f is a homeomor- phism. Proof of (16.9). Suppose that f has the lifting property. Pick y 2 Y and let V be a simply connected neighbourhood of y.
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